Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61
Answer:
The tension in the rope at the lowest point is 270 N
Explanation:
Given;
weight of the ball, W = 150 N
length of the rope, r = 4 m
velocity of the ball, v = 5.6 m/s
When the ball passes through the lowest point, the tension on the rope is the sum of weight of the ball and centripetal force.
T = W + F
Centripetal force, F = mv²/r
where;
m is the mass of the ball
m = W/g
m = 150 / 9.8 = 15.306 kg
Centripetal force, F = mv²/r
F = (15.306 x 5.6²)/4
F = 120 N
T = W + F
T = 150 + 120
T = 270 N
Therefore, the tension in the rope at the lowest point is 270 N
Answer:
Explanation:
ADT for an 2-D array:
struct array{
int arr[10];
}arrmain[10];
An application that stores an array with 1000 rows and 1000 columns, where less than 10,000 of the array values are non-zero. The two different implementations for such arrays that would be more space efficient than a standard two-dimensional array implementation requiring one million positions are :
1) struct array{
int *p;
}arr[1000];
2) struct array{
int *p;
}arr[1000];
Answer:
if you are speaking of the acronym then Engineering uses science and mathematics to solve everyday problems in society
Answer:
The pressure drop is 269.7N/m^2
Explanation:
∆P = ∆h × rho × g
∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2
∆P = 0.032×860×9.8 = 269.7N/m^2
